Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
in [Relativity]
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From: Thomas Smid on 12 Sep 2005 10:51 Daryl McCullough wrote: > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> As I have told you several times, x2 is not a variable. It is > >> (by assumption) equal to -ct. x1 is not a variable. It is equal > >> to +ct. So you can't go from > >> > >> ct'(x1,t) = (B+A) x1 > >> to > >> ct'(-x1,t) = -(B+A)x1 > >> > >> That amounts to going from > >> > >> ct'(ct,t) = (B+A) ct > >> > >> to > >> > >> ct'(-ct,t) = - (B+A) ct > >> > >> That's clearly incorrect, and you keep doing it. > > > >We have gone through all this already. Each time the algebra is not > >working in your sense, > > Each time, you fail to keep track of dependencies between > variables, and it gets you into trouble. As I have said > before, if you want to show that *I'm* making a mistake, > then you need to be *more* careful than I am. But you are > consistently *less* careful. > > >you are trying to change the formalism by adding > >arguments to the variables. > > I'm saying that that is the *correct* way to reason > about these things. If you want to show *my* reasoning > leads to a contradiction, then you need to use *my* > reasoning. If you want to show that *your* reasoning > leads to a contradiction, then by all means, use your > own reasoning. > > You are the one who introduced x1 and said that it was > equal to ct. You are the one who introduced x2 and said > that it was equal to -ct. Thus it makes no sense to go > from > > >> ct'(x1,t) = (B+A) x1 > >> to > >> ct'(-x1,t) = -(B+A)x1 > > How can it make any sense, when x1 is just another > name for ct? > > >This does not change anything about the > >algebraic conclusions > > The correct algebraic conclusion is that you are having > trouble with algebra. Have you ever heard of *checking* > your results? You say that the equations > > 1. forall e, x'(e) = A x(e) + Bc t(e) > 2. forall e, ct'(e) = D x(e) + Ec t(e) > 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > > imply that A=0 or B=0. Obviously, that's incorrect, because > one solution to these 4 equations is this: A = 5/3, B = -4/3 > > x' = 5/3 x - 4/3 ct > ct' = -4/3 x + 5/3 ct > > Let's check: Suppose x=ct. Then x' = 5/3 ct - 4/3 ct = 1/3 ct > Then t' = -4/3 ct + 5/3 ct = 1/3 ct. So x' = ct'. Check. > > Suppose x = -ct. Then x' = -5/3 ct - 4/3 ct = -3 ct. > Then t' = +4/3 ct + 5/3 ct = +3 ct. So x' = -ct'. Check. > > So, obviously my 4 equations above are consistent with both > A and B being nonzero. So if you come up with A=0 or B=0, > then you are adding something new. If that something new > leads to a contradiction, then that's *your* contradiction, > not mine. You haven't actually solved your equations (1)-(4) as they contain D and E which you replaced by A and B (and it is here where the contradiction occurs (as I mentioned before)) > > >Assume you have a function f(x) which is defined for all numbers x > >(positive as well as negative) and you have the two equations > > > >(1) f(x)=(A+B)x > >(2) f(-x)=(A-B)x > >then you can *logically* conclude from (1) that > >(3) f(-x)=-(A+B)x > >and thus by comparison with (2) > >(4) A=0. > > Right. But we are not in that situation. We have a function > of *two* variables x and t. > > >In the same sense, if your function t'(x,t) is defined on all numbers x > >(positive and negative) and you have the relationships > > > >(5) ct'(x1,t)=(B+A)x1 > > But that equation is *not* true for all x and t. It is only > true when x = ct. > > >(6) ct'(x2,t)=(B-A)x2 > > But that equation is *not* true for all x and t. It is only > true when x = -ct. > These are conditions you interprete into the validity of the variables which aren't fixed mathematically anywhere. There is only one function t'(x,t) defined here i.e. if you are serious about this issue you should also split t into t1 and t2, but I wonder how far you will be getting this way in the derivation of the Lorentz transformation. Also you still haven't answered my question why you are not changing the sign convention and let x1=ct and x2=ct. Thomas
From: Daryl McCullough on 12 Sep 2005 11:24 Thomas Smid says... > >Daryl McCullough wrote: >> 1. forall e, x'(e) = A x(e) + Bc t(e) >> 2. forall e, ct'(e) = D x(e) + Ec t(e) >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) >> >> imply that A=0 or B=0. Obviously, that's incorrect, because >> one solution to these 4 equations is this: A = 5/3, B = -4/3 >> >> x' = 5/3 x - 4/3 ct >> ct' = -4/3 x + 5/3 ct >You haven't actually solved your equations (1)-(4) as they contain D >and E which you replaced by A and B (and it is here where the >contradiction occurs (as I mentioned before)) I'm sorry. I thought it was clear what D and E were, but to be explicit: The 4 equations above have a solution A=5/3 B=-4/3 D=-4/3 E=5/3 So it is incorrect to say that the 4 equations imply A=0 or B=0. There is no contradiction. >> >(5) ct'(x1,t)=(B+A)x1 >> >> But that equation is *not* true for all x and t. It is only >> true when x = ct. >> >> >(6) ct'(x2,t)=(B-A)x2 >> >> But that equation is *not* true for all x and t. It is only >> true when x = -ct. > >These are conditions you interprete into the validity of the variables >which aren't fixed mathematically anywhere. I don't know what you are talking about. As I said, these are the four conditions we have talked about so far: >> 1. forall e, x'(e) = A x(e) + Bc t(e) >> 2. forall e, ct'(e) = D x(e) + Ec t(e) >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) Equations 1 and 2 say "forall e". Equations 3 and 4 have conditions. You are ignoring these conditions. >There is only one function >t'(x,t) defined here If you don't like events, then the same information can be conveyed in terms of x and t: 1. forall x, forall t, x'(x,t) = A x + B c t 2. forall x, forall t, t'(x,t) = D x + E c t 3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t) 4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t) >i.e. if you are serious about this issue you >should also split t into t1 and t2, but I wonder how far you will be >getting this way in the derivation of the Lorentz transformation. > >Also you still haven't answered my question why you are not changing >the sign convention and let x1=ct and x2=ct. You were the one who said that x2 was -ct. Now you want to say that x2 = +ct. That doesn't make any sense. -- Daryl McCullough Ithaca, NY
From: Dirk Van de moortel on 12 Sep 2005 12:54 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:dg46ij014k1(a)drn.newsguy.com... > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> 1. forall e, x'(e) = A x(e) + Bc t(e) > >> 2. forall e, ct'(e) = D x(e) + Ec t(e) > >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > >> > >> imply that A=0 or B=0. Obviously, that's incorrect, because > >> one solution to these 4 equations is this: A = 5/3, B = -4/3 > >> > >> x' = 5/3 x - 4/3 ct > >> ct' = -4/3 x + 5/3 ct > > >You haven't actually solved your equations (1)-(4) as they contain D > >and E which you replaced by A and B (and it is here where the > >contradiction occurs (as I mentioned before)) > > I'm sorry. I thought it was clear what D and E were, but to be > explicit: The 4 equations above have a solution > > A=5/3 > > B=-4/3 > > D=-4/3 > > E=5/3 > > So it is incorrect to say that the 4 equations imply A=0 or B=0. > There is no contradiction. > > >> >(5) ct'(x1,t)=(B+A)x1 > >> > >> But that equation is *not* true for all x and t. It is only > >> true when x = ct. > >> > >> >(6) ct'(x2,t)=(B-A)x2 > >> > >> But that equation is *not* true for all x and t. It is only > >> true when x = -ct. > > > >These are conditions you interprete into the validity of the variables > >which aren't fixed mathematically anywhere. > > I don't know what you are talking about. > > As I said, these are > the four conditions we have talked about so far: > > >> 1. forall e, x'(e) = A x(e) + Bc t(e) > >> 2. forall e, ct'(e) = D x(e) + Ec t(e) > >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > > Equations 1 and 2 say "forall e". Equations 3 and 4 > have conditions. You are ignoring these conditions. > > >There is only one function > >t'(x,t) defined here > > If you don't like events, then the same information can > be conveyed in terms of x and t: > > 1. forall x, forall t, x'(x,t) = A x + B c t > 2. forall x, forall t, t'(x,t) = D x + E c t > 3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t) > 4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t) > > >i.e. if you are serious about this issue you > >should also split t into t1 and t2, but I wonder how far you will be > >getting this way in the derivation of the Lorentz transformation. > > > >Also you still haven't answered my question why you are not changing > >the sign convention and let x1=ct and x2=ct. > > You were the one who said that x2 was -ct. Now you want to say > that x2 = +ct. That doesn't make any sense. Daryl, even sub-retards like Marcel Luttgens and Androcles would laugh their bottoms off over this Smid character. Have you considered asking him to solve an equation like 4x - 3 = 5 ? Are you still not convinced that he is playing a game? Dirk Vdm
From: Thomas Smid on 12 Sep 2005 14:01 Daryl McCullough wrote: > Thomas Smid says... > > > >Daryl McCullough wrote: > > >> 1. forall e, x'(e) = A x(e) + Bc t(e) > >> 2. forall e, ct'(e) = D x(e) + Ec t(e) > >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > >> > >> imply that A=0 or B=0. Obviously, that's incorrect, because > >> one solution to these 4 equations is this: A = 5/3, B = -4/3 > >> > >> x' = 5/3 x - 4/3 ct > >> ct' = -4/3 x + 5/3 ct > > >You haven't actually solved your equations (1)-(4) as they contain D > >and E which you replaced by A and B (and it is here where the > >contradiction occurs (as I mentioned before)) > > I'm sorry. I thought it was clear what D and E were, but to be > explicit: The 4 equations above have a solution > > A=5/3 > > B=-4/3 > > D=-4/3 > > E=5/3 > > So it is incorrect to say that the 4 equations imply A=0 or B=0. > There is no contradiction. No there isn't a contradiction as you restricted again the validity of the equations: if you have two arbitrary events e1 and e2 , the transformation equations are x'(e1) = A x(e1) + Bc t(e1) ct'(e1)= B x(e1) + Ac t(e1) x'(e2) = A x(e2) + Bc t(e2) ct'(e2)= B x(e2) + Ac t(e2) These equations must hold for all events e1 and e2 simultaneously, whether x(e1) and x(e2) are positive or negative, and thus also for further events at -x(e1) and -x(e2), so we can actually double up the equations to (1a) x'(e1) = A x(e1) + Bc t(e1) (1b) ct'(e1)= B x(e1) + Ac t(e1) (2a) -x'(e1) = -A x(e1) + Bc t(e1) (2b) ct'(e1)= -B x(e1) + Ac t(e1) (3a) x'(e2) = A x(e2) + Bc t(e2) (3b) ct'(e2)= B x(e2) + Ac t(e2) (4a) -x'(e2) = -A x(e2) + Bc t(e2) (4b) ct'(e2)= -B x(e2) + Ac t(e2) Note that at this point e1 and e2 are arbitray and we have not defined a relationship x(e2)=-x(e1) yet. Still it is obvious from (1b) and (2b) for instance that B=0. > > >> >(5) ct'(x1,t)=(B+A)x1 > >> > >> But that equation is *not* true for all x and t. It is only > >> true when x = ct. > >> > >> >(6) ct'(x2,t)=(B-A)x2 > >> > >> But that equation is *not* true for all x and t. It is only > >> true when x = -ct. > > > >These are conditions you interprete into the validity of the variables > >which aren't fixed mathematically anywhere. > > I don't know what you are talking about. > > As I said, these are > the four conditions we have talked about so far: > > >> 1. forall e, x'(e) = A x(e) + Bc t(e) > >> 2. forall e, ct'(e) = D x(e) + Ec t(e) > >> 3. forall e, if x(e) = ct(e), then x'(e) = ct'(e) > >> 4. forall e, if x(e) = -ct(e), then x'(e) = -ct'(e) > > Equations 1 and 2 say "forall e". Equations 3 and 4 > have conditions. You are ignoring these conditions. > > >There is only one function > >t'(x,t) defined here > > If you don't like events, then the same information can > be conveyed in terms of x and t: > > 1. forall x, forall t, x'(x,t) = A x + B c t > 2. forall x, forall t, t'(x,t) = D x + E c t > 3. forall x, forall t, if x=ct, then x'(x,t) = c t'(x,t) > 4. forall x, forall t, if x=-ct, then x'(x,t) = -c t'(x,t) > > >i.e. if you are serious about this issue you > >should also split t into t1 and t2, but I wonder how far you will be > >getting this way in the derivation of the Lorentz transformation. > > > >Also you still haven't answered my question why you are not changing > >the sign convention and let x1=ct and x2=ct. > > You were the one who said that x2 was -ct. Now you want to say > that x2 = +ct. That doesn't make any sense. On the contrary, it makes perfect sense if you have separate variables for the two directions. You only need negative numbers if you have one variable. Our calendar is for instance the best example for this: you have one scale named AD and another named BC, both measured positive. Obviously you could do the same for a spatial scale as well. If you want, you can define the location of the light signal by a spherical coordinate e.g. by r(phi,t) (where phi is a corresponding angle and thus r always positive). I am sure you are not suggesting that the derivation of the Lorentz transformation depends on the use of a Cartesian coordinate system. Thomas
From: Dirk Van de moortel on 12 Sep 2005 14:28
"Thomas Smid" <thomas.smid(a)gmail.com> wrote in message news:1126548118.449286.86870(a)g14g2000cwa.googlegroups.com... [snip] > I am sure you are not suggesting that the derivation of the Lorentz > transformation depends on the use of a Cartesian coordinate system. It depends on linear algebra, which is something you seem to pretend never to have heard about. So either you are lying about this, or you are lying about having a masters in physics. Or you have severe brain rot. What will it be? Dirk Vdm |